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Calculus Level 3

$\displaystyle\sum _{ n=0 }^{ \infty }{ \frac { { e }^{ n } }{ n! } }$

The value of the above expression is in the form $A^B$ , then find $\ln(AB)$ .

1 solution

Akshay Yadav
Feb 17, 2016

Note that,

$e^x = 1+ \frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!} ...$

And the series given to us is,

$\displaystyle\sum _{ n=0 }^{ \infty }{ \frac { { e }^{ n } }{ n! } }= 1+\frac{e}{1!}+\frac{e^2}{2!}+\frac{e^3}{3!} ...$

Observe the similarity in both the series.

Hence,

$\displaystyle\sum _{ n=0 }^{ \infty }{ \frac { { e }^{ n } }{ n! } }=e^e$

$A= e$ and $B= e$

$\ln(AB)=\ln(e.e)=\ln(e^2)$

$2\ln(e)$

$\boxed{2}$

exactly as intended!

Hamza A - 5 years, 3 months ago

You are missing one term !!

$\frac{e^{0}}{0!}=1$

Akshat Sharda - 5 years, 3 months ago

Thanks Akshat, I have made the necessary changes.

Akshay Yadav - 5 years, 3 months ago